CHAPTER X. SIMULTANEOUS EQUATIONS.

CHAPTER X. SIMULTANEOUS EQUATIONS. 140. A SINGLE equation which contains two or more unknown quantities can be satisfied by an indefinite number of values of the unknown quantities. For we can give any values whatever to all but one of the unknown quantities, and we shall then have an equation to determine the remaining unknown quantity. If there are two equations containing two unknown quantities (or as many equations as there are unknown quantities), each equation taken by itself can be satisfied in an indefinite number of ways, but this is not the case when both (or all) the equations are to be satisfied by the same values of the unknown quantities. Two or more equations which are to be satisfied by the same values of the unknown quantities contained in them are called a system of simultaneous equations. The degree of an equation which contains the unknown quantities x, y, z... is the degree of that term which is of the highest dimensions in x, y, z Thus the equations ax + d 2 y + a z z = a 4, xy + x + y + z = 0, a?+y 2 + z 2 - Sxyz = 0, are of the first, second and third degrees respectively.

144 SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 141. Equations of the First Degree. We proceed to consider equations of the first degree, beginning with those which contain only two unknown quantities, x and y. Every equation of the first degree in x, y, z,... can by transformation be reduced to the form ax + by + oz -f... = k, where a y b, c,... k are supposed to represent known quantities. NOTE. When there are several equations of the same type it is convenient and usual to employ the same letters in all, but with marks of distinction for the different equations. Thus we use a, b, c... for one equation; a', b\ c... for a second ; a", b"> c"... for a third ; and so on. Or we use a lt b lf c x for one equation; a 2,b 2, c 2 for a second; and so on. Hence two equations containing x and y are in their most general forms ax+hy=c } and a'x + b'y = c, and similarly in other cases. 142. Equations with two unknown quantities. Suppose that we have the two equations ax+by = c, and a x -f b'y = c'. Multiply both members of the first equation by b\ the coefficient of y in the second; and multiply both members of the second equation by b, the coefficient of y in the first. We thus obtain the equivalent system ab'x + bb r y = cb', abx 4- bb'y = cb.

SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 145 whence Hence, by subtraction, we have (ab' db) x = cb'' c'b; cb' c'b x = ab' db Substitute this value of x in the first of the given equations; then cb' cb 7 ab ab, _ c (ab' a'b) a (cb' cb), ac' a'c whence y = y-,, f. u ab ab The value of y may be found independently of x by multiplying the first equation by a and the second by a \ we thus obtain the equivalent system dax 4- a'by = a'c, a'ax + ab'y ac. Hence, by subtraction, we have (a'b ab f ) y = dc ac'; '' _ ac ac' y ~a'b~^ab" which is equal to the value of y obtained by substitution. NOTE. It is important to notice that when the value either of x or of y is obtained, the value of the other can be written down. For a and d have the same relation to x that b and V have to 3/; we may therefore change x into y provided that we at the same time change a into b, b into a, a into b\ and V into d. Thus from cb' cb, cd ca x = 77 n we nave y= x f yj-. ab ab ba ba S. A. 10

14G SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. It will be seen from the above that in order to solve two simultaneous equations of the first degree, we first deduce from the given equations a third equation which contains only one of the unknown quantities; and the unknown quantity which is absent is said to have been eliminated. 143. From the last article it will be seen that the values of x and y which satisfy the equations ax + by = c, and a'x + h'y = c', can be expressed in the form y -1 be b'e ca ca cb f a'b * So also, from the equations ax -f by + c = 0, and ax + b'y + c = 0,. x y 1 we have be b'c ca ca ab' a'b* It is important that the student should be able to quote these formulae. Ex. 1. and We have Solve the equations 3.r + 2^ = 13, 7a: + 3z/=27. y -L 2. 27-3.13 "13. 7-27. 3 3.3-7.2' that is x _ y 1 # 15""l0'~5 ; and y = = 2.

SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 147 Ex. 2. Solve the equations x y A 2 5 and = - 7. x y These may be considered as two simultaneous equations of the first degree with - and - as unknown quantities. x y "We therefore have 1 x 1 y -1»(- -7)-(-5)2 "2.2 -(-7)4- = 4(-~5) - 2. 3 ' 1 1 that is X y i. "32 26' Also 1 x~~ 1 y~ 11 26' 32 ""26' or x=- 26 11* 13 o r^ = 16 Ex. 3. Solve the equations x-y = a-b t ax-by = 2(a*-b 2 ), We have X y -1-2i{a*-b 2 ) + b(a -b) a («- b) - 2 {a*-6«) " -6 + a' that is and ' X y 1 b* + ab- 2a 2 2b 2 -ab- ar ~~ 6- a' b 2 + ab- 2a 2 _. x= b + 2a; b-a 2b 2 -ab -a 2 _ a + 2b. Instead of referring to the general formulae of Art. 143, as we have done in the above examples, the unknown quantities may be eliminated in turn, as in Art. 142; and this latter method is frequently the* simpler of the two. Thus in this last example we have at once, by multiplying the first equation by a and then subtracting the second, (b - a) y = a (a - b) - 2 (a 2 - b 2 ); -a 2 -ab + 2b 2 y = i = a + 2b. b-a 10 2

148 SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. Then x = {a + 2b)+a- b;.-. x = 2a + b. 144 Discussion of solution of two simultaneous equations of the first degree. We have seen that the values of x and y which satisfy the equations ax + by =c and ax + b'y c (ii), are given by (ab' - a'b) x = cb' cb (bd b'a) y = ca c'a (i), (iii), (iv). Thus there is a single finite value of x, aud a single finite value of y, provided that ab' a'b =f= 0. If ab' ab = 0, x will be infinite [see Art. 118] unless cb r c'b = Q; and, if ab' a f b and cb' c'b are both zero, any value of x will satisfy equation (iii). So also, y will be infinite if ab' a'b = 0, unless cd c'a is also zero, in which case any value of y will satisfy equation (iv). If ab' a'b = 0, then = j,; and if ab' a'b=0 a o and also cb' c'b = 0, then = = -,. a b c When equations cannot be satisfied by finite values of the unknown quantities, they are often said to be inconsistent. Thus the equations ax-^-by c and a'x + b'y = c are inconsistent if > = 77 > unless each fraction is equal to a b ^ -j, in which case the equations are indeterminate. c In fact when = 77 =, it is clear that by multiplying the terms of equation (i) by we shall obtain equation (ii), so that a the two given equations are equivalent to one only.

SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 149 We have hitherto supposed that a, a, b, b' were none of them zero. It will not be necessary to discuss every possible case: consider, for example, the case in which a and a! are both zero. When a and a are both zero, we have from (i) y = j, and from (ii) y jj. These results are inconsistent with one another unless T = o b r.. c c' Hence, if a = a = 0, and T=T,> the equations (i) and c (ii) are satisfied by making y = T> and by giving to x any finite value whatever. c c' If however j- =j= j-,, the equations by = c and b'y = c cannot both be satisfied, unless they are looked upon as the limiting forms of the equations cuc+by=c and dcc+b'y=c', in which a and a are indefinitely small and ultimately zero. But from (iii) we see that when a and a! diminish without limit, x must increase without limit, cb' cb not being zero. Thus, in the equations (i) and (ii), when a and a' diminish without limit, and cb' 4= c% the value of x must be infinite. Equations with three unknown quantities. 145. To solve the three equations: ax + by + cz = d (i), dx-^b'y-^c'z d' (ii), a"x+b"y + c"z = d" (iii). Method of successive elimination. Multiply the first equation by c', and the second by c; then we have acx + bey -f ccz = dc\

150 SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. and a ex -4- b'cy -f e'ez d'e ; therefore, by subtraction, (ac a'c) x -t- (6c' b'c) y = dc d'e (iv). Again, by multiplying the first equation by c" and the third by c and subtracting, we have (ac" - d'e) x + (be" - b"c) y = dc" - d"c (v). We now have the two equations (iv) and (v) from which to determine the unknown quantities x and y. Using the general formulae of Art. 143, we have ^ - (fr c ' - y c ) (dc" - d"c) + (dc - d'e) (be" - b"c) (ac a'c)(be" b"c) (be b'c) (ac" a'c) Method of undetermined multipliers. Multiply the equations (i) and (ii) by X and //,, and add to (iii); then we have the equation x (Xa + pa + a") + y(\b + fib' + b") + z (Xc + fie' + c") = (Xd + fid' + d"), which is true for all values of X and p. Now let X and fi be so chosen that the co-efficients of y and z may both be zero, Xd + ud' + d" then x = ^ -, T,, Xa + /xa + a where X and ft are found from X6+ A i6' + 6" = 0, and Xc + /*c' + c" = 0; _ V* ' 6 V - ft'v " 6"c - 5c" be - b'c ' Hence _ d (Vc" - 6V) + d' (b"c - 6c") + d" (be' - 6'c) X ~~ a (b'c" - b"c) + a! [b"c - 6c") + a" (be' - 6'c)' [The numerator and the denominator of the first value of x, which was obtained by eliminating z and y in succes-

SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 151 sion, can both be divided by c; and the two values of x will then be seen to agree.] Having found the value of x by either of the above methods, the values of y and z can be written down. For the value of y will be obtained from that of x by interchanging a and 6, a' and b\ and a" and b". The value of y can also be obtained from that of x by a cyclical change [see Art. 93] of the letters a, b, c; a, b\ c ; and a", b"> c"; and a second cyclical change will give the value of z. It should be remarked that the denominators of the values of x y y and z are the same, and that there is a single finite value of each of the unknown quantities unless this denominator is zero. Ex. 1. Solve the equations : x + 2y + 3z = (i), 2x + 4y + z = l (ii), Sx + 2y + 9z = 14 (iii). Multiply (ii) by 3, and subtract (i); then 5x + 10y = 15 (iv). Again multiply (i) by 3, and subtract (iii); then 4^ = 4 (v). From (v) we have y = 1; then, knowing y, we have from (iv) x = l; and, knowing x and?/, we have from (i) z = 1. Thus x=y =2 = 1. Ex. 2. Solve the equations : x+y+z=l (i), ax + by + cz = d (ii), a?x + b 2 y + c 2 z = d 2 (iii). Multiply (i) by c and subtract (ii); then (c - a) x + (c - b) y = c - d (iv). Again multiply (i) by c 2 and subtract (iii); then {cz-a 2 )x + (c*-b2)y = c 2 -d* (v). Now multiply (iv) by c + b and subtract (v); then (c - a) (b- a)x = (c - d) (b - d); _{b-d)(c-d) *' (b-a)(c -a)'

152 SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. The values of y and z may now be written down: they are (c-d)(a-d) m (a-d)(b-d) J (c-b)(a-b) ' {a-c)(b-c) ' Instead of going through the process of elimination, we may at once quote the general formulae. Thus _ {be 2 - b 2 c) + d {b 2 - c 2 ) + d 2 (c - b) X ~{bc*-b 2 c) + a{b 2 -c 2 ) + a 2 {c-b) _(b- c){ - be + d{b + c) - d 2 } ~~ (b-c){ -bc + a(b + c)-ar\ (b -d)(e-d), = 77 T7 f» as above. (b - a) (c - a) Ex. 3. Solve the equations: x + y + z--a + b-t-c (i), ax + by + cz = be + ca + ab (ii), hex + cay + abz = Sabc (iii). We have _ (a -f b + c) {ab 2 - ac 2 ) + {be + ca + ab) {ca - ab) + 3abc {c - b) ~ ab 2 -ac 2 + a{ca-ab) + bc{c-b) a{b - c) {{b + c) {a + b + c) - be - ca - ab - Sbc} ~~ (6-c) [ab \-ae-a 2 -bc} a{b-c) 2 ~~ {a-b)(a c)' The values of y and z can now be written down: they are h ^c ~ a)2 c ( a ~ h f V ~ (b-c)(b-a) ; Z ~ (c-a)(c-b)' Ex. 4. Solve the equations : x+ ay + a 2 z + a* = 0 (i), x + by+b 2 z + b* = 0 (ii), x + cy + c 2 z + c 3 =0 (iii). The equations may be solved as in the preceding examples, or as follows. It is clear that a, 6, c are the three roots of the following cubic in X \* + z\ 2 + y\ + x = 0. Hence from Art. 129, we have at once z= ~{a + b + c), y = bc + ca+ab y and x = - abc.

SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. 153 146. Equations with more than three unknown quantities. We shall return to the consideration of simultaneous equations of the first degree in the Chapter on Determinants, and shall then shew how the solution of any number of such equations can be at once written down. The method of successive elimination or the method of undetermined multipliers can however be extended to the case when there are more than three unknown quantities. For example, to solve the equations ax + by + cz +dw =e (i), ax + b'y + c'z +d'w = e' (ii), d'x + b"y + c'z +d"w =e" (iii), a"'* + b'"y + d"z + d'"w = e" (iv). Multiply (i) by X, (ii) by fi, (iii) by v, and add the products to (iv). Then we have x (ax + a'fi + a"v + a"') + y (b\ + ftp + b"v + b'") -f z(c\ + c'fm + c"v + c") + w(d\ + d'fi + d"v + d'") = e\ + e'fm + e"v + e" (v). Now choose X, ji, v so as to make the coefficients of y, z and w in the last equation zero; then e\ + e u, 4- e v + e,. x oo= - j- //7 (vi), where X, fi, v are to be found from the equations b\ + b'p + b"v + V" = 0^ cx + c'fm +c"v +c'" = 0> (vii). Hence we have to solve (vii) by Art. 145 and then substitute the values of X, fi, and v in (vi) ; this will give the value of x; and the values of the other unknown quantities can then be found by cyclical changes of the letters, a, b> c, d y &c.

154 SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. EXAMPLES XIL Solve the following equations. 1 x y \ L. 3 6 2> x 3y 1 5 " 10 ~ 2 ' 3. 3 7 2 y o 2 26 2/ 3 0 4 2.?-* = 2, J 4 3 r 6 3 4. +5=-+-= 10. as y a y 5. ax + by = 2ab, 6. a; + a?/ + a 2 = 0, bx-ay = b 2 - a 2. x + by + b 2 = 0. 7. x + y = 2a, 8. (5 + c) x + (b-c)y = 2ab t (a b)x=(a + b)y. (c + a) x +(c-a)y = 2ac. 9. bx + ay = 2ab y a 2 x + 6 2 y = a 3 + b 3. 10. (a + b)x-\-by = ax+ (6 + a)y = a 3 b 3, 11. x + y + z-l 9 12. x + y + z = l, 2x+3y + z = 4, x y,.. 4a+9?/ + s=lg. 2 i + ' 5 3 z, in 13. a; + 2y + 32; = 3a: + y + 2z = 2x + 3y + z = 6. 14. y + z=2a, 15. y + 3- c = 2a, z + x=2b, z + x-y=2b, x + y=2c. x + y-z = 2c,

EXAMPLES. 155 16. y + z~ 3x = 2a, % + x - 3y = 26, x + 2/ - 3s = 2c. 17. ax + by + cz = 1, 6cc + cy + az = 1, cx + ay + bz 1.?/ + z -x _% + x y x + y-z 18. *.~_r- = 64-c c 4-a a + 6 = 1. 19. 21 x + y + z =, 20. ax 4- by 4- cz = 1, era? 4-6 2 y 4- c 2 s = a 4-6 4- c. SC4-2/4- = a4-&4-c, bx + cy + az = a 2 + b 2 + c 8, ex 4- ay 4-6s = a 2 + 6 2 4- c 2. x4-2/4-s = a4-64-c, 6x + cy + az = bc +ca + ab y ex + ay + bz = bc + ca+ab. 22. 23. 25. 26. 27. 28.,T42/42 = 0, (b +c) x + (c + a)y + (a + b)z = (b c) (c - a)(a- 6), bcx + cay + abz = 0. ax + by 4- cz = a, bx + cy + as = 6, cx + ay 4-6s = c. ax + by ± cz = m, a 9 x 4-6 2 y.4- c 2 s = m 2, a 3 x 4- b 3 y 4- c 3 s = m 3. ax + cy + bz = a 2 + 26c, cx + by+az = b 2 + 2ca, bx + ay + cz = c* + 2ab. x + y + z = 2a + 2b + 2c, ax + by 4- cs = 26c 4-2ca 4-2a6, (6 c) x 4- (c a) y + (a 6) z = 0. ax 4-6y 4- cs = a + 6 4- c, a 2 x + b 2 y + c 2 z = (a + 6 4- c) 2, 6cx 4- cay + abz = 0. 24. x - ay + a 2 z - a 3 = 0, x-6y+6 2 s-6 3 =0, as cy 4- c 2 s c 3 = 0.

156 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. 29. x+y + z = l + m + n, Ix + my -r nz = mn + nl + lm } (m n) x + (n - I) y + (I - m) z = 0. 30. Ix + ny + mz - nx + my + lz = mx + ly + nz = l 3 + m 8 + n 3 - Zlmn. 31. l 2 x 4- m 2 y + n 2 z = Imx + mny + nlz = nix 4- Imy + mnz = 1+ m + n. x y z 32. + r>+ =1, a + a a + p a + y x y z b + a 6 + /3 b + y~ ' +» + = l - C + a c + p c 4- y 33. y + z + w = a, z + w-hx = b, w-t-x + y = c, x + y + z = d. 34. x 4- ay + a 2 z + a 3 w + a* = 0, x + by + b 2 z + b 3 w + b* = 0, a; 4- cy + c 2 z 4- c 8 w; 4- c 4 = 0, x+dy + d 2 z + d 3 w 4- cz 4 = 0. Simultaneous Equations of the Second Degree. 147. We now proceed to consider simultaneous equations, one at least of which is of the second or of higher degree. We first take the case of two equations containing two unknown quantities, one of the equations being of the first degree and the other of the second.

SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. 157 For example, to solve the equations : Sx + 2y = 7, 3ar 2-2y 2 = 25. From the first equation we have x- 3. Substitute this value of x in the second equation; we then have whence y 2 +Uy + 13 = -0, that is (y + 13)(y + l) = 0; If y=-- 1,.?/= -1, or y-= - 13..-^-., and if y= -13, «= 11. Thus x=s, y = 1; or # = 11, y= -13. From the above example it will be seen that to solve two equations of which one is of the first degree, and the other of the second degree, we proceed as follows: From the equation of the first degree find the value of one of the unknown quantities in terms of the other unknown quantity and the known quantities, and substitute this value in the equation of the second degree ; one of the unknown quantities is thus eliminated, and a quadratic equation is obtained the roots of which are the values of the unknown quantity which is retained. The most general forms of two equations such as we are now considering are Ix + my + n = 0, ax 2 + Ixy + cy 2 + dx + ey +/= 0. From the first equation we have my + n x- l.

158 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. Hence on substitution in the second equation we have to determine y from the quadratic equation a (my + n) 2 Iby (my + n) + cl 2 y 2 - dl {my + n) + epy +fl 2 = 0. Having found the two values of y, the corresponding values of x are found by substitution in the first equation. 148. It should be remarked that we cannot solve any two equations which are both of the second degree; for the elimination of one of the unknown quantities will in general lead to an equation of the fourth degree, from which the remaining unknown quantity would have to be found; and we cannot solve an equation of higher degree than the second, except in very special cases. For example, to solve the equations aa? + bx + c = y, x 2 + y 2 =d. Substitute ax 2 + bx + c fory in the second equation, and we have x 2 + (ax 2 + bx + cf = d } which is an equation of the fourth degree which cannot be solved by any methods given in the previous chapter. 149. There is one important class of equations with two unknown quantities which can always be solved, namely, equations in which all the terms which.contain the unknown quantities are of the second degree. The most general forms of two such equations are ax 2 + bxy + cy 2 = d ' and ax 2 + b'xy + cy 2 = d'. Multiply the first equation by d', and the second by d and subtract; we then have (ad' - ad) x 2 + (fid 1 - b'd)xy + (cd'- cd) y 2 = 0.

SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. 159 The factors of the above equation can be found either by inspection, or as in Art. 81; we therefore have two equations of the form Ix + tny = 0 either of which combined with the first of the given equations will give, as in Art. 147, two pairs of values of x and y. Ex. 5. To solve the equations: y 2 -xy = W (i), x 2 + xy = 14 (ii). We have 14 {y 2 - xy) = 15 (x 2 + xy);.*. 15x 2 + 29^'-14?/- = 0, that is {5x - 2y) {Sx + 7y) = 0. Hence ox-2y = 0, or else Sx + ly = 0. If 5x - 2y = 0, we have from (i) 2T--2/- = 15, whence y = ± 5. Hence also x = ± 2. If Sx + 7y 0, we have from (i) 3 whence y= : ± s/2 7 and then x= : "V2 7 3 ; or 1 = Thus #=±2, y=±o; or a?=±-^, y=^f -75. 150. The following examples will shew how to deal with some other cases of simultaneous equations with two unknown quantities; but no general rules can be given. Ex.1. To solve x-y = 2, xy = 15. 7 = 15, Square the members of the first equation, and add four times the second; then (a; +?/) 2 = 64.

160 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. Hence x + y=±8 y which with x - y = 2, gives x = 5 or -3, and y = 3 or -5. Thus x = 5, y = 3; orx=-3,y=-5. Ex.2. To solve x* + xy+y~ = a- (i), x* + xy + y*=b i (ii). Divide the members of the second equation by the corresponding members of the first; then b* x 2 -xy + y 2 = (iii). From (i) and (iii) by subtraction we have From (i) and (iv) b* 2xy = a' 2 x 2 + 2xy + y 2 = 2 ; /3a*-b*., a+ y =sfc V-2Si- w- From (iii) and (iv) we have 2 _ t. 3b*-a* x 2-2xy + y 2 = 2a2 ~; (iv). ' ar -y =± \/-25r- Finally, from (v) and (vi) we have and _ 1 J 4- /3<* 4 -& 4, /3b*-a*) X ~2 V 2d> V 2 «2 f ' 1 ( /3a*-b* /3b* ~ a*) 2/ = 2 J* sj-%*- T \/~^M " Ex. 3. To solve a: 2-2y 2 = 4y, 3x 2 + x?/-2t/ 2 =16y. Multiply the first equation by 4, and subtract the second; then x 2 -xy-&y 2 = 0, that is (x + 2y) (x - 3y) = 0; /. x + 2y = 0, or else a; - 3y = 0. ^

SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. 161 If x + 2y = 0, the first equation gives.". V = 0 or y = 2, whence x = 0 or x= -4. If x - % = 0, the first equation gives 9y 2-2y*=4y; 4 /. y = 0 or y- at = 7 r whence x = 0 or x = 12 = 7 ' Thus x = 0, y=0; x = = -4, y = i or 12 4 Ex. 4. To i?olve * 2 +?/ 2 = (;z + t,-fl) 2, By subtraction we have tf 2 +?/ 2 =(a;- y + 2)\ (a?+y + l) 2 -(jj--y + 2) 2 = 0, that is (2* + 3)(2y--1) = 0. Hence If 2* + 3: = 0, we i have whence 2a;+3 = 0, or! 2 y-l=0. +2/ 2 =(2/- i)'- y=-2 If 2y -1 = 0, we have ^H*^) 2 ' whence 2 3 Thus ^ " ^ y~~ 2; 2 1 Ex.5. To solve x+y = 26, #< + i/ 4 =2a 4. Put x=b+z; then, from the tirst equation, y = 6-«. S. A. 11

162 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. Hence (6 + s) 4 + (b - z)* = 2a\ whence after reduction * 4 + 6& 2 2 2 + & 4 -a 4 =0;.-. a2=-36 2 ± N /86 4 + a 4 ; Thus x = b±j{-zb*±jwt^}, and y = &T > /{-36 2 ± N /86 4 + a 4 }. EXAMPLES XIII. Solve the following equations : 1. x + y = x* y 2 = 23. % x 2-4y 2 + x+3y = 2x-y=l. 3. a: 2 + a^=12, 4. x>+2y 2 = 22, xy-2y 2 =l. 3y*-xy-x*=U. 5. x - y ~ 5, 6. x + y = a + 6, a b x + ~b + ; y +a = 1. 1 _! - A 2/ x 84 * 7. a(x + y) = b (x y) = xy. o 1 1 1 8. -j + =, 9. x xy a a 2 a; 2 ab 10, y* yx~ b 2 ' 10. a; + y= 2a, 11. a; 3 + y 3 = 26 3. xy~ 3. a; 2 --xy + y 2 = 109, x 4 + x 2 y* + jf = -4251, 12. x 2 + xy + y 2 = 133, 13. a; + y = 72, a?+ V«y + y = 19. %x+ %y=6.

EXAMPLES. 163 14. 1+1=2, 15. <e+y=l, xy + - + - = 8. x 5 + w 5 = 31, J x y 16. a; 2 + y 2 + 3^- 4(a? + y) + 3 = 0, scy + 2 (# +?/) - 5 = 0. 17. a; 3 + #«/ + # =14, 18. x 3 + y 3 = 9, y 2 + xy + y= 28. x*-xy + y 2 = 3. 19. x(y-b) =y(x-a) = 2ab. 20. x + - = l, y i, y + i = 4. 21. ax+by= 2ab, 22. 2 2 1 1 1 x y 3. # 3 23. + #y = a 2,? +xy = b*. 24. xy-^ = a, 25. x + y +??- = 14, y i xy- = -. * cc a ^ + ^ + ^ = 84. 26. a + y = 6, 27. x + y= 8xy, (x 2 + y 2 ) (a 3 + 2/ 3 ) = 1440. a 2 + y» = 40ccy. 28. x*-xy = 8x+3 f 29. ~~-^ = 3, * ' 1 - xy ' &y - y* - 8y - 6. a?/ 1 1 + icy 3 * 11 2

164 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. 30. x-y = a(x 2 -y 2 ), x + y = b(x 2 -y 2 ). nr% x a y b x y ax b y y x 31. x y b a.f ~ = a b a b ' x z y 2 b 2 a 9 a 1 + b 2 ~a 2 + b 2 ' 151. Equations with more than two unknown quantities. No general rules can be given for the solution of simultaneous equations of the second degree with more than two unknown quantities: all that can be done is to solve some typical examples. Ex. 1. Solve the equations: (x + y)(x + z)=a* (i), {y + z)(y + x) = b 2 (ii), (z+x)(z+y)=c* (iii). Multiply (ii) and (iii) and divide by (i); then 6 2 c 2 (2/+*) 3 =-^-; Similarly we have " y+z== 7 < lv) * 2+ * T (v) ', ab... and x + y= ± (vi). Also from the original equations it is clear that the signs must all be positive or all be negative. Add (v) and (vi) and subtract (iv) from the sum; then. (ca ab bc\ \b c a J cw + aw-bw *"* 2abc So also y=± aw + bw-cw ^ 2 2abc a u 9 and b*c* + c 2 a?-a?b 2 z = i 2abc

SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. 165 Ex. 2. Solve the equations: x(y+z)=a y{z + x)=b z(x + y)=c (i), (ii), (iii). We have y(z + x) + z(x + y) -x{y + z) = b + c-a t that is 2yz = b + c-a. Similarly 2zx=e + a-b t and 2xy a + b-c. Hence ( 2ay) {2zx) = ( a + b " c )( c + a - }) ). 2yz b + c-a ' 2x2 = (a + b-c)(c + a-b) ^ (b + c-a) 6-c) V 2(b-fc-a] 0 ' and similarly y = ± / ^ 0,,. v. J * \/ 2(c + a-5) and ^^^e+c^ohc he-a) (c + a o-6) 2(a + b-c) Ex. 3. Solve the equations: x 2 + 2yz=a y 2 + 2zx a z 2 +2xy = b (i), (ii), (iii). By addition (x + y + z) 2 = 2a + b ;.\ x + y + z= ±j2a + b (iv). From (i) and (ii) by subtraction (x-y)(x + y-2z) = 0. Hence x=y (v), or else x + y-2z-0 (vi). I. If x=y y we have from (ii) and (iii) by subtraction z 2 -f x 2-2xz = 6 - a; /. z-x ±.*Jb-a (vii). Hence, from (iv), (v) and (vii),

166 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. II. When x + y - 2z 0, we have from (iv) z=±-j2a + b, and x + y = Also, from (ii), y 2 + x (x + y) = a, which with the previous equation gives V a-b 1, and y = =F K/^Y~ ± s <j2a + b. Ex. 4. Solve the equations: fc 2 2 + c 2^ = c 2 c 4- a 2 z = a 2 y + b 2 x = a-?/;?. We have b 2 z + chj = xyz... (i), c 2 x + a 2 z = xyz (ii), and a 2 y + b 2 x=xyz (iii). Multiply (i) by - a 2, (ii) by 6 2, and (iii) by c 2, and add; then 2b 2 c 2 x =(-a* + b 2 + c 2 ) xyz. Hence 0 r e l s e x=0 f 2b 2 c 2 V Z = 2, 1.8, 2 If a;=0,?/ and 2 must also be zero. Hence x=y=z~q; or else yz - 2b 2 c 2 b 2 + c 2 n.2' and similarly z*= ci+ C J_ bi» 2a 2 6 2 and xy = - 'a 2 + b 2 -c 2 ' The solution then proceeds as in Ex. 2. Ex. 5. Solve the equations: x 2 -yz a, y 2 -zx = b, z 2 -xy=c.

SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. 167 We have (x 2 - yz) 2 (y 2 - zx) (z* - xy) = a 2 ~ be, that is x (x? + y* + 2 s - Sxyz) = a 2 - be. Hence, from the last equation and the two similar ones, & V _ z a 1 -be b 2 - ca c' 2 -ab' Hence each fraction is equal to / *?-y* _ ± / i V (<>? ~ *>c) 2 - {b 2 - ca) (c 2 - ab)~ \ (a 3 + 6 3 + c 3 - Sabc) ' Ex. 6. Solve the equations: x+y+z=a+b+c x 2 + y 2 + z*=a? + b' 2 + c 2 (i), (ii),?+*+*=3 (iii). a b c ' It is obvious that x = a, y b, z c will satisfy the equations: put then x=a + \, y = b + fjt., z = c + v, and we have after reduction X + /i + p=--0 (iv), ^ + + ^=0 (v), a b c v J 2(a\ + bijl + cp) +\ 2 + /jt 2 + v 2 = 0 (vi) From (iv) and (v) X _ fi _ v a(b - c) ~ b(e ~ a) ~~ c (a - b)' whence from (vi) X 2(b-e)(c-a){a-h) a(b-c)~ a 2 {b - c)' 2 + b 2 (c - a) 2 + c 2 (a-b) 2 ' Hence x = a, y = b, z c\ or else x-a y-b z-c 2{b-c)(e-a) (a-b) cr(f-c) ~ b(c-a) ~" c~(a - b) ~ a 2 (b - cf + b 2 {c - af + c 2 (a- bf' Ex. 7. Solve the equations: x + y+z= 6, yz 4- zx + xy = 11, xyz= 6. This is an example of a system of three symmetrical equations. Such equations can generally be easily solved by making use of the relations of Art. 129. Thus in the present instance it is clear that x t y, z are the three roots of the cubic equation \ 3-6AHlX\-G = Q.

168 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. The roots of the cubic are 1, 2, 3. Hence x=l, # = 2, z=3; or JC=1, y = 3, z = 2; or x = 2, T/ = 3, JB = 1; &c. Ex. 8. Solve the equations: x + y + z = a (i), 1 1 1 1 -+-+-=- (n), x y z a ' yz + zx+xy= -c 2 (iii). This again is a system of symmetrical equations, and two of the relations of Art. 129 are already given; we have therefore only to find the third. We have from (ii), yz + zx + xy _1 1 xyz ~ a'.*. xyz= -ac 2 (iv). Then, from (i), (iii) and (iv), we see that x, y, z are the roots of the cubic X s - a\ 2 - c 2 \ + ac 2 = 0, that is X 2 (X - a) - c 2 (X - a) = 0;.*. X=a, or X= ±c. Thus x=a, y=c, z = c; &c. Ex. 9. Solve the equations: x 2 (y-z) = a 2 (h-c), y 2 (z - x) = b 2 (c - a), z 2 (x-y) = c 2 (a-b). By addition x 2 {y-z) + y 2 (z-x)+z 2 {x-y) = a?(b-c) + b 2 {c-a) + c 2 {a-h), that is (y -z)(z- x) (x-y) = (b-c)(c -a)(a- b). By multiplication x 2 y 2 z 2 (y - z) (z - x)(x - y) = d 2 b 2 c 2 {b -c){c- a) (a-b);.\ x 2 y 2 z 2 =a 2 &c 2. Hence xyz = abe (i), or xyz= -abc (ii). Again a 2 (b - c)y + b 2 (e - a)x=x 2 y{y -z) + xy' 2 (z - x) =xyz(y-x) (iii). Hence, if xyz abc, we have from (iii) {b 2 (c-a) + abc] x-\- {a?(b-c) - abc}y = 0, that is bx (be + ca- ab) ay (be + ca- ab) = 0 ;.". - = ;-, and therefore each = -. a b c

EXAMPLES. 169 Thus, when xyz=abc, we have - = f = -. a b c Hence each is equal to / -y- = ^/l. Thus - = ^ = -=1, or = -^- = = 1, a o c a<a bo) aa x y _ z aw 2 ~" bco 2 ~ cu? "" If xyz = ahc t we have from (iii) - (be-ca ab)=j (ca-ah-he). Hence also each = - (ah - he - ca) =\/{ -(hc ca ah) (ca- ah-he) (ab hc ca)}. EXAMPLES XIV. Solve the following equations : 1. yz = a 2, 2. x (x + y + z) = a*, zx = b 2, y (x + y + z) - b 2 > xy = c 2. z (x + y + z) = c 2. 3. yz + zx + xy = a, 4. yz = a (y + z), yz zx + xy = b t zx = b(z + x), yz + zx xy = c. xy = c (x + y). 5. yz by + cz, 6. x 2 + 2yz = 12, zx = cz + aa?, 2/ 8 + 2zx = 12, xy = ax + by. z* + 2xy = 12. (y + z)(x + y + z) = a, 8. (y + 6) (z + c) = a* (z + x) (x + y + z) = b, (z + c) (x + a) = 6 2 (^+y) ( x + 2/ + z ) = c - ( x + a ) (y + fy=c 2.

170 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE. 9. x 2 -(y-z) 2 = a 2, 10. x (y + z - x) = a, y 2 -(z- xf = b*, y(z+x-y)=b, z 2 -(x- y) 2 = <?. z (x+ y z) = c. y + z z + x x + y 11. ^ = j- = Z = 2xyz. a b c * y + z z + x x+y x 2 + y* + z 9 12. a b c a 2 + b 2 + c z ' 13. yz = a + y + z, 14. yz a (y + z) + a, 2# = 6 + z + x, zx = a(z + x) + fi, xy = c + x + y. xy = a(x + y) + y. 15. yz -f 2 = cy + bz, 16. zx g 2 az + ex, xy h 2 = bx + ay. 17. x <r+ + <*/ y + z * = c, G, 18. x 2 + y 2 + z 2-14, xyz = 6. 19. # x + ^ y + 2= z = 9, 20. a?+2/ + s= 10, x 2 + y* + z 2 = 41, y«+ zx + xy = 33, x 3 + y* + z 3 = 189. (y+*)(*+a>)(a;+y) = 294. 21. x + y l 3 ~~2' y + z- 1 7 ~3' z+x~ l = 4. x + y ' + 3 = 15, ** + y 3 + z 3 = 495, xyz 105. y*l- bz + cy zx a?/ x 2 + y 2 + z* ex + az ay + bx a 2 + b 2 + c*' a y z 22. -+ +- = X b c i, 1, 23. ax = ^- + - y z y X b z T z x + - + - = a i, by = -+-, y e x % X y c a b z = i. x y cz = - +-. y «

24. y 2 + z*-x(y + z) = a 2, z 2 + x 2 -y(z + x) = b 2, x 2 + y 2 -z(x + y) = c 2. EXAMPLES. 171 25. x 2 + yz-a 2 = y 2 + zx- 6* =z 2 + xy - c" = ^ (x 2 4- y 2 + z 2 ). 26. x (x + y + z) - (y 2 + z 2 + yz) = a, y (x + y + z) - (z 2 4- x 2 + zx) = 6, z (x + y + z) - (x 2 + y 2 + xy) = c. 27. aj + 2/ + «= a4-6+c, a? 8 4- y 2 + z 2 = a 2 + b 2 + c a, (b c) x 4- (c - a) 3/ 4- (a b) z = 0. 28. (a: 4-y) (a: 4-2) = a#, 29. x 2 yz = ax, {y + z ) (y + x ) = h> y*-zx = by, (z + x)(z +y) = cz. z 2 - xy = cz. 30. x 2 +a(2x + y + z)=y 2 + b(2y+z + x) = z 2 + c(2z + x + y) = (a; + y + z) 2. 31. y 2 + yz + z 2 = a\ z 2 + zx + x* = b 2 > x* + xy + y 2 = c 2. 32. a 2 x + b 2 y + c 2 z = 0, (b-c) 2 {(c-ay l(a-h)^0 ax by cz 1 1 1 1 1 1 4-4- =-+- + -. yz zx xy a o c f